1. Text Exercise 1.22 (a) (b) (c) (d) (This part is asking you whether the data looks reasonably bell-shaped.) The distribution looks somewhat close to bell-shaped. y =s = 0.8425 0.345525 0.8425  2(0.345525) 0.15145, 1.53355 With a bell-shaped distribution, we expect this percentage to be about 95%. The bars on the histogram represent the intervals 0 to 0.15, 0.15 to 0.30, …, 1.80 to 1.95. The first bar and the last three appear to be mostly outside the interval 0.15145 to 1.53355. Therefore, the actual percentage inside the interval appears to be about 92% or 93%. Homework #2Score____________/ 15Name ______________

2. Text Exercise 1.24 - continued (b) Use the SPSS output produced in part (a) in answering this part of the exercise, and be sure to explain the reasoning behind your answer thoroughly. Step 4:Make certain that no item in the output is selected. Use the File> Print options to obtain a printed copy of the histogram and the mean and standard deviation which appear to the lower right of the histogram. Since there is no need for you to save the output, you may close the SPSS output window without saving the results, after you have your printed copy of the output. Attach this printed copy to this assignment before submission. Exit from SPSS. y =s =from the SPSS output. 48.16 6.015 Since the histogram looks reasonably bell-shaped, we can be about 95% certain that a randomly selected age from the women on the list will be in the interval 48.16  2(6.015), that is, between 36.13 and 60.19.

3. Text Exercise 1.26 (a) (b) (This part is asking you whether the data looks reasonably bell-shaped.) (Make use of the second statement of Tchebysheff’s Theorem in answering this part.) The distribution looks very positively skewed and not at all bell-shaped. At least 89% of the measurements must lie within three standard deviations of the mean. We can use this to predict with at least 89% confidence that the next major oil spillage amount will lie between 59.8 – 3(53.36) and 59.8 + 3(53.36) metric tons. Since oil spillage amount must be positive, we can change this prediction interval to between 0 and 219.88 metric tons.

4. Additional HW Exercise #1.2 (a) (b) Consider the following data set:50 55 58 59 60 61 62 65 70 Find the mean, the variance, and the standard deviation. Indicate how one value in the data set could be changed to make all of the values in the data set except one larger than the mean. y =s 2 =s =60 32.55.70 Change the smallest value 50 to any value smaller than 5.

5. Additional HW Exercise #1.3 In a sample of 25 male grackles, each at least one year old, the wing chords were measured in centimeters and recorded as follows: 14.114.013.914.413.013.614.313.613.8 14.113.714.314.213.414.113.514.814.4 13.814.013.514.513.513.814.4 The stem-and-leaf displays below use stems 13.-, 13.t, 13.f, 13.s, 13.*, 14.-, etc., where “-” represents 0s and 1s, “t” represents 2s and 3s, “f” represents 4s and 5s, “s” represents 6s and 7s, and “*” represents 8s and 9s. 13.- 13.t 13.f 13.s 13.* 14.- 14.t 14.f 14.s 14.* 10 9 4 0 6 3 6 8 1 7 32 4 1 5 8 4 8 0 5 5 5 8 4 13.- 13.t 13.f 13.s 13.* 14.- 14.t 14.f 14.s 14.* 00 8 4 0 6 2 6 8 1 7 33 4 1 5 8 4 8 1 5 4 5 9 5

6. Notice that the stem-and-leaf display on the left contains only the first row of data (i.e., the first 9 observations). Complete the stem-and-leaf display on the left, and complete the construction of the ordered stem-and-leaf display on the right. Find the five-number summary, the median, the range, and the interquartile range. Additional HW Exercise #1.3 – continued (a) (b) Min= Q 1 = Q 2 = Q 3 = Max= 13.0 14.8 14.0 13.6 14.3 = median IQR =14.3 – 13.6 = 0.7 range =14.8 – 13.0 = 1.8

7. Decide whether or not there are any potential outliers, and construct a box plot. (c) 12.813.213.614.014.4 Wing Chords (centimeters) 14.813.013.413.814.214.615.0 1.5(0.7) = 1.05 13.6 – 13.0 = 0.6 < 1.05 14.8 – 14.3 = 0.5 < 1.05 No outliers